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Q)

$A 12 \;pF$ capacitor is connected to a $50\;V$ battery. How much electrostatic energy is stored in the capacitor?

$\begin{array}{1 1} 0.05 \times 10^{-8} \\ 2.5 \times 10^{-8}\\ 0.5 \times 10^{-8} \\ 1.5 \times 10^{-8}\end{array} $

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A)
Solution :
Capacitor of the capacitance, $C = 12\; pF = 12 \times 10^{−12} F$
Potential difference, V = 50 V
Electrostatic energy stored in the capacitor is given by the relation,
$e= \large\frac{1}{2} $$CV^2 =\frac{1}{2} \times 12 \times 10^{-12} \times (50)^2J = 1.5 \times 10^{-8}\;J$
Therefore, the electrostatic energy stored in the capacitor is $1.5 \times 10^{-8}\; J$
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